Is this matrix $M^{-1}A$ which often appears in finite element method for parabolic PDEs diagonalizable? Both are N-by-N and tridiagonal and diagonal dominant matrices.
$$M=\begin{pmatrix} 4 & 1 & & & \\ 1 & 4 & 1 & & \\ & 1 & \ddots & \ddots & \\ & & \ddots & 4 & 1 \\ & & & 1 & 4 \end{pmatrix}$$
$$A=\begin{pmatrix} - & + & & & \\ + & - & + & & \\ & + & \ddots & \ddots & \\ & & \ddots & - & + \\ & & & + & - \end{pmatrix}$$
With experiments with randomly generated matrix A, it seems that it's right. But can anyone give some clues for rigorous proof?
Thanks!
Since $M$ is SPD and $A$ is symmetric, $M$ has an SPD square root $M^{1/2}$ (with the SPD inverse $M^{-1/2}$). The matrix $M^{-1}A$ is similar to $M^{1/2}(M^{-1}A)M^{-1/2}=M^{-1/2}AM^{-1/2}$, which (since $A$ and $M^{-1/2}$ are symmetric) is symmetric and hence diagonalizable.
The case when $A$ is not symmetric is tricky. For example, $A$ being diagonalizable is not a sufficient condition for $M^{-1}A$ being diagonalizable (and vice versa). For example, $$ M=\begin{bmatrix}4&1\\1&4\end{bmatrix}, \quad A=\begin{bmatrix}16&8\\4&17\end{bmatrix}. $$ Here $M$ is SPD and $A$ is diagonalizable (in addition, $A$ is diagonally dominant). However, $$ M^{-1}A=\begin{bmatrix}4 & 1\\ 0 & 4\end{bmatrix} $$ is not diagonalizable.
Note that since the set of diagonalizable matrices is a dense subset of $\mathbb{R}^{n\times n}$ (w.r.t. any norm), a random matrix (without a special structure) is almost sure to be diagonalizable.